Primitive roots are useful in analyzing the multiplicative structure of the numbers mod $p$, for $p$ a prime. I wondered if there was anything that could link the multiplicative and additive structure together. I thus came up with the following conjecture:
For all sufficiently large primes $p$, there are two different primitive roots mod $p$ adding to a third.
Given the large proportion of primitive roots a number usually has, this seems quite plausible. A preliminary computer search shows that the conjecture holds for $p = 11$ and $17 \le p \le 1{,}000$. The primes $p \in \{2, 3, 5, 7, 13\}$ are counterexamples.
This solution is due to a colleague, which is why I'm marking it as a community wiki. Surprisingly, we're able to prove the bound $p \ge 17$ directly.
We first prove that mod $p \ge 17$, there's a sixth power $a$ and a square $b$, neither in $\{0, 1\}$, with difference $1$. Let $g$ be a primitive root mod $p$ such that $g^{12} \ne 2$, which exists as either a primitive root or its reciprocal satisfies this. Note also that $g^6 \ne \pm 1$. If either of $g^6 \pm 1$ is a quadratic residue, we're done. Otherwise, $g^{12}$ and $(g^6 - 1)(g^6 + 1) = g^{12} - 1$ satisfy the desired property.
Now, for $g$ a primitive root mod $p$, write $a = g^{6m}$, $b = g^{2n}$. Note that among the numbers $\{6m, 2n, 0\}$, only one congruence mod $2$ and at most two congruences mod $3$ are represented. By the Chinese residue theorem, we can find some $k \ne 0$ such that for any prime $q$ coprime with $p - 1$, $$k \not\equiv -6m,\,-2n,\,0 \pmod q.$$ Then the numbers $g^{6m+k}$, $g^{2n+k}$, $g^k$ are distinct primitive roots, and two of them add to the third.